Moments Of Inertia

[gunterZA]

Hi,

I need some assistance understanding moments of inertia. I am doing some review for an upcoming exam, however I am slightly stumped by this question. I have already tried googling "moments of inertia" to try and understand the concept better, but I am having issues knowing when to apply which formulas.

I have attached my specific question and the "answer" to the question, hopefully someone can help me understand how to get the answer.

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Two identical slender rods of length l and mass m are linked together at 90deg, as shown in figure Q4a (image attached), to form a link in a mechanism. Point B is midway between A and C, and points B, D, E, F and G are equally spaced along the lower link.

i) Determine the position of the center of gravity of the link

ii) Find the moment of inertia of the link about the point B

iii) Find the moment of inertia of the link about the point D

iv) Find the radius of gyration about the point G




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ANSWERS:

i) Center of gravity is at D

ii) I_B=(5/12)ml² kg m²

iii) I_D=(7/24)ml² kg m²

iv) k= sqrt(17/24) l m

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*My attempts*


i)


I know inherently that the centre of gravity is at D. How do i prove it mathematically?

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ii)


I get the moment of inertial about B like so:

I_B = (ml²)/3 + ((ml²/3) - (ml²)/4)

But i don't understand why I am supposed to subtract. Is it because the centre of gravity is l/2 below B?

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iii)


For I_D I get the answer by doing:

I_D = ((ml²)/3 - (ml²)/4) + ((ml²)/(3*16)) + (9ml²)/(16*3)

Why do I subtract (ml²)/4) if that rod is above D?


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iv)


I'm trying to use I_G to find the radius of gyration:

So for I_G

I_G = (7/24)ml² - ml²


Here I am using the parallel axis theorem, but why do I subtract the (md²) of the AC rod from the Moment of inertia about D?
 

[kabhay]

As for first part choose point B as origin. Now you know the co-ordinate of centre of mass of rod AC and the co-ordinate of centre of mass of rod BG. Now apply the formula of centre of mass.

[OceanBreeze]

To solve for the centroid, you can pick any point on the object to be the reference axis, but it helps to pick a convenient point. In this case, point G is the most convenient as the entire T can be rotated about it in one plane. Point B would require rotations in two planes.

 

The centroid X_G  = M1X1 + M2X2 / M1 + M2

 

We don’t know the masses but that is not a problem because it is a uniform rod, the length can stand in for the mass.

 

X_G = L1X1 + L2X2 / L1 + L2

X_G = (4x4) + (4x2) / 4 + 4 = 3

So, the centroid is located a distance of 3 units from point G, placing it at point D.

 

Solving for the Moment of Inertia, (I) is a bit more complicated but still very easy.

 

The basic formula is dI = X² dm , and since dm = M/L dx , dI = M/L X² dx

 

I = M/L ∫ X² dx

 

When integrated between the limits L-h and h, where h is some arbitrary distance along the length,

 

I = 1/3 M [ L² -3Lh + 3h² ]

When h = 0 as in the vertical part of the T shape, I = 1/3 ML²

When h = L/2, as in the case of the horizontal part of the T , I = 1/12 ML²

The total moment of Inertia of the T  around the point B is just the sum of the two parts, 5/12 ML²

 

The rest is left as an exercise for the readers